$$yu_x-xu_y+u_z=1$$
The change to cylindrical coordinates would simplify the calculus. Nevertheless, we will solve the PDE in Cartesian coo1dinates, just to show that it is not very difficult.
The differential characteristic equations are :
$\quad \frac{dx}{y}= \frac{dy}{-x}=\frac{dz}{1}=\frac{du}{1}$
A first family of characteristic curves comes from $dz=du \quad\to\quad u-z=c_1$
A second family of characteristic curves comes from :$\quad\frac{dx}{y}= \frac{dy}{-x} \quad\to\quad x^2+y^2=c_2$
A third family is a bit more difficult to find : $ \frac{dz}{1}=\frac{dx}{\sqrt{c_2-x^2}} \quad\to\quad \tan^{-1}\left(\frac{x}{\sqrt{c_2-x^2}} \right)-z=c_3 $
On the characteristic curves $c_1,c_2,c_3$ are independent. Elsewhere they are related by an implicit equation :
$$\Phi\left(u-z\:,\:x^2+y^2\:,\:\tan^{-1}\left(\frac{x}{y} \right)-z\right)=0$$
where $\Phi$ is any differentiable function of three variables.
An equivalant manner to express the relationship is :
$$u-z=F\left(x^2+y^2\:,\:\tan^{-1}\left(\frac{x}{y} \right)-z\right)$$
where $F(X,Y)$ is any differentiable function of two variables.
This is the general solution of the PDE in Cartesian coordinates :
$$u(x,y,z)=z+F\left(x^2+y^2\:,\:\tan^{-1}\left(\frac{x}{y} \right)-z\right)$$
Or in cylindrical coordinates :
$$u=z+F\left(\rho^2\:,\:\tan^{-1}\left(\frac{\cos(\theta)}{\sin(\theta)} \right)-z\right) =z+F\left(\rho^2\:,\:\frac{\pi}{2}-\theta-z\right) $$
Now, we consider the boundary condition :
$$u(x,y,0)=x+y=\rho\left(\cos(\theta)+\sin(\theta)\right)=F\left(\rho^2\:,\:\frac{\pi}{2}-\theta\right)$$
Thus, the function $F(X,Y)$ is determined :
$$F(X,Y)=\sqrt{X}\left(\cos(\frac{\pi}{2}-Y)+\sin(\frac{\pi}{2}-Y) \right)=\sqrt{X}\left(\sin(Y)+\cos(Y) \right)$$
Bringing it back into the above general solution, with $X=\sqrt{x^2+y^2}$ and $Y=\tan^{-1}\left(\frac{x}{y} \right)-z$ , the particular solution of the PDE according to the boundary condition is :
$$u(x,y,z)=z+\sqrt{x^2+y^2}\left[\sin\left(\tan^{-1}\left(\frac{x}{y} \right)-z \right) +\cos \left(\tan^{-1}\left(\frac{x}{y} \right)-z \right) \right]$$
So you have
$$
\frac{dx}{-z/p^2}=\frac{dt}{1}=\frac{dz}{q-z/p}=-\frac{dp}{1}=-\frac{dq}{q/p}
$$
leading directly to $$p+t=p_0,~ x+q=x_0+q_0,~ q/p=q_0/p_0,~ z/p^2=z_0/p_0^2,$$ and then in combination $x+z_0/p_0^2t=x_0$.
The PDE at $t=0$ gives $q_0+z_0/p_0=0$. The initial condition evaluates to $z_0=x_0^2/2$, $p_0=u_x(x_0,0)=x_0$, $q_0=-z_0/p_0=-x_0/2$. This simplifies the equations for the characteristic so far to
$$
p+t=x_0,~q+x=\tfrac12x_0,~ q/p=-\tfrac12,~z/p^2=\tfrac12,~x+\tfrac12t=x_0
$$
The solution tangent plane equation gives
$$
dz = p\,dx+q\,dt = -\tfrac12p\,dt-\tfrac12p\,dt=-(x_0-t)\,dt
\\~\\
z=z_0-x_0t+\tfrac12t^2=\tfrac12(x_0-t)^2=\tfrac12(x-\tfrac12t)^2
$$
Often one can condense down such exercise solutions to a much narrower set of identities.
Best Answer
Multiplying the first equation by $x$ and the second one by $y$ gives $$ \tfrac12 (x^2)' = -\frac{x^2 z}{x^2 + y^2}, \qquad \tfrac12 (y^2)' = -\frac{y^2 z}{x^2 + y^2} . $$ The sum leads to $(r^2)' = -2z$ where $r^2 = x^2 + y^2$. Multiplying the first equation by $y$ and the second one by $x$ gives $$ yx' = -\frac{xy z}{x^2 + y^2}, \qquad xy' = -\frac{xy z}{x^2 + y^2} . $$ The difference leads to $t' = 0$ where $t = y/x$. The additional equations $u' = 0$ and $z' = 1$ yield $u = c_1$ and $z = s + c_2$, respectively. Thus, integrating the differential equations for $r^2$ and $t$, we find $r^2 = -s(s + 2 c_2) + c_3$ and $t = c_4$. The fact that $z^2 = -r^2+ c_3 + (c_2)^2$ leads to a general solution of the form $u = F(x^2+y^2+z^2, y/x)$ for some arbitrary $F$.
The above steps suggest that the cylindrical coordinates $(x,y,z) \mapsto (r, \theta, z)$ may be relevant here. Indeed, the PDE rewrites as $zu_r - r u_z = 0$, which general solution is $u = G(r^2 + z^2, \theta)$ for some arbitrary $G$.